Prove that $|f(\frac{p}{q})|\geq \frac{1}{q^4}$ where $p,q$ are integers Let $f(x)\equiv n_0x^4+n_1x^3+n_2x^2+n_3x+n_4=0$,where $n_0,n_1,n_2,n_3,n_4$ are integers.If $f(x)=0$ has two distinct irrational roots whose product is also irrational then prove that $|f(\frac{p}{q})|\geq \frac{1}{q^4}$ where $p,q$ are integers.
In this question,i tried to apply rational root theorem,but not applicable.What is the condition for two irrational roots?Please help me.
 A: This is more a comment than an answer.
This is the start
of how Liouville constructed
the first transcendental numbers.
The basis of this construction
is the following theorem:
For a polynomial 
$f(x) \in \mathbb{Z}[x]$ 
of degree $n$, 
we have the following result:
If $p, q \in \mathbb{Z}, q \ne 0$
(assume $q \ge 1$, safe) 
and $f(p/q) \ne 0$,
then
$|f(p/q)|
\ge 1/q^n
$.
A nice proof is here:
http://deanlm.com/transcendental/construction_of_a_transcendental_number.pdf
Your problem
is the case $n=4$.
From this you can readily
show that
if $x$ is an irrational root
of 
$f(x) \in \mathbb{Z}[x]$ 
of degree $n$,
then there is a constant $M$
defined in terms
of the coefficients
of $f(x)$
such that
$|\frac{p}{q}-x|
\ge \frac{1}{Mq^n} 
$.
This theorem says that
an irrational algebraic number of
degree $n$ cannot be approximated
too well by a rational number.
What Liouville did
was to construct a number $x$
that could be so accurately
approximated by a sequence of rational numbers
that
for any $M$,
for all large enough $n$,
there are
$p, q \in \mathbb{Z}$
such that
$|\frac{p}{q}-x|
< \frac{1}{Mq^n} 
$.
This means that $x$
can not be
algebraic of any degree
and so must be transcendental.
Please read that article for all the details.
